If NASA faked the moon landings, does the agency have any credibility at all? Was the Space Shuttle program also a hoax? Is the International Space Station another one? Do not dismiss these hypotheses offhand. Check out our wider NASA research and make up your own mind about it all.

Maybe it's me, but I don't see any real benefit in debunking any YouTube video, apart from spending hours. If you haven't read the Almagest, in the first pages there is an explanation of why the Earth's shape is not plane: "if it were plane, the stars would rise and set simultaneously for everyone on Earth".

brianv » 25 Nov 2017, 14:41 wrote:Great job debunking. Even I nearly understood. So what is the calculated distance to the Sun using trig with these values? Which I think was the object of the exercise.

I remember you weren't interested in any YouTube video or in mathematics. Have you changed your mind?

brianv » 25 Nov 2017, 14:41 wrote:Great job debunking. Even I nearly understood. So what is the calculated distance to the Sun using trig with these values? Which I think was the object of the exercise.

I remember you weren't interested in any YouTube video or in mathematics. Have you changed your mind?

Oh, occasionally I suppose if I think I might learn something...there are levels of interested. There are videos I wouldn't touch with a barge-pole and there are those which pop up on my "recommended panel", thanks to the videos that are posted here. But, I mostly use YouTube for music theory and production and my eldest loves 70's UK Cult Kids TV like Timeslip, Children of the Stones, Tomorrow People etc etc...all on YouTube, and I have to admit a certain nostalgic fondness too. Not to mention that a friend of mine "starred" in one of them!

Don't remember ever saying I'm not interested in Maths, I do have a Degree in Computer Science you know. I daresay few here know or understand what Orthogonal is.

In mathematics, orthogonality is the generalization of the notion of perpendicularity to the linear algebra of bilinear forms. Two elements u and v of a vector space with bilinear form B are orthogonal when B(u,v) = 0. When the bilinear form corresponds to a pseudo-Euclidean space, there are non-perpendicular vectors that are hyperbolic-orthogonal. In the case of function spaces, families of orthogonal functions are used to form a basis.

I wouldn't have known either except for Blender's Perspective and Orthographic Views, if I'm not mistaken.

brianv » November 25th, 2017, 6:41 am wrote:Great job debunking. Even I nearly understood. So what is the calculated distance to the Sun using trig with these values? Which I think was the object of the exercise.

You can't really calculate the sun's distance accurately with those values due to the sun position calculator having an accuracy of 0.5 arcminutes (30 arcseconds or 0.0083°), and calculating the sun's distance from those values would require the angles be accurate within 1 or 2 arcseconds.

With that limitation in mind, we can come up with a possible range of values for the sun's distance using the data for locations identified in the video as "B" and "F" because they are very close to the same distance apart from "D". Be warned, maths ahead.

At B the sun's elevation is 47.0832°. At F it is at 47.8468°.We adjust by 0.3817 to account for the sun's altitude at D (89.6183°) being offset by 0.3817 from 90°, so add 0.3817 to B and subtract it from F (because these locations are in opposite directions from D):B = 47.0832 + 0.3817 = 47.4649°F = 47.8468 - 0.3817 = 47.4651°

Now all we have to do is multiply the B-F chord length (straight-line distance) by the cosecant of the difference between the two angles. If we use those angles as-is without accounting for the ± 0.0083 uncertainty, the difference is 0.0002°. Given that the B-F chord length is 8613 km, that equals:8613*csc(0.0002°) = 2,467,442,744 kmBut we know that distance isn't accurate because it's based on an angle far smaller than the stated accuracy of the data.

If we account for the ± 0.0083 uncertainty, the difference between angles B and F could be as high as 0.0168°, which corresponds to a distance of:8613*csc(0.0168°) = 29,374,318 km. This would be the lower bound for distance based on the uncertainty of the angles - the sun can be further away than that, but it cannot be closer (closer would require a larger angular difference, and we can't get that even with the uncertainty).

Hmm. Not being smart or anything, but that leaves us back where we started. The video-maker couldn't measure the distance using the wrong values, and RP can't calculate with the real values. Doesn't the NOAA Calculator take the arc minutes and seconds into account? Isn't this a case of the "real angles" versus "perceived angles"?

How can this be correct if the distance to the Sun can't be measured?

Andromeda

Messier 31 or M31 (also designated NGC 224) is a spiral galaxy in the constellation Andromeda. It has an apparent visual magnitude of 3.4 and its angular diameter is 178x63 arc-minutes. M31 lies at an estimated distance of 2.9 million light years. The Equinox 2000 coordinates are RA= 0h 41.8m, Dec= +41° 16´

Amongst the lay public of non-mathematicians and non-scientists, trigonometry is known chiefly for its application to measurement problems, yet is also often used in ways that are far more subtle, such as its place in the theory of music;

The sun's distance can be measured, just not from the values used in the video. Measuring distance from elevation angles also requires that the angle difference between the local horizons be calculated with arcsecond precision, and I doubt that the 360° * (distance/circumference) achieves that level of precision.

For example, when Venus is as maximum elongation (furthest angular separation from the sun) you can compare two different images of it taken at the same time from two locations sufficiently far (a few hundred kilometers at least) apart on earth.

By noting how far it's angular separation from a background star changes (parallax) between the two images, along with the distance between the two observing locations where the images were taken, you can calculate it's distance using trigonometry.

Now you can use the distance to Venus as one leg of a right triangle, along with Venus' angular separation from the sun, and determine the sun's distance with trigonometry as well.